bemath-If-each-point-on-the-line-3x-4y-2-is-transformed-by-matrix-M-2-0-0-1-the-image-is-a-line-

Question Number 110948 by bemath last updated on 01/Sep/20

\sqrt{bemath}

If each point on the line 3 x + 4 y = 2

is transformed by matrix M = (\begin{matrix} 2 0 \\ 0 1 \end{matrix}), the

image is a line___

Commented by bemath last updated on 01/Sep/20

hshahaha \dots . bemath sir . bemath macho

Commented by Rasheed.Sindhi last updated on 01/Sep/20

T h a n k y o u m y d e a r!

Commented by bobhans last updated on 01/Sep/20

standart way

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} 2 0 \\ 0 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) \Rightarrow (\begin{matrix} x \\ y \end{matrix}) = \frac{1}{2} (\begin{matrix} 1 0 \\ 0 2 \end{matrix}) (\begin{matrix} x^{'} \\ y^{'} \end{matrix})

(\begin{matrix} x \\ y \end{matrix}) = \frac{1}{2} (\begin{matrix} x^{'} \\ {2 y}^{'} \end{matrix}) = (\begin{matrix} \frac{1}{2} x^{'} \\ y^{'} \end{matrix})

hence the image of line is \frac{3}{2} x^{'} + {4 y}^{'} = 2

multiply by 2 \Rightarrow 3 x + 8 y - 4 = 0

Commented by Rasheed.Sindhi last updated on 01/Sep/20

\sqrt{bemath} = \sqrt{b} \sqrt{e} \sqrt{m} \sqrt{a} \sqrt{t} \sqrt{h} :)

Commented by bemath last updated on 01/Sep/20

��

Commented by Rasheed.Sindhi last updated on 01/Sep/20

BTW, b e - m a t h o r b e m - a t h

o r s o m e t h i n g e l s e ?

Answered by bobhans last updated on 01/Sep/20

say the image is a line ℓ \equiv x | \begin{matrix} 3 4 \\ 0 1 \end{matrix} | + y | \begin{matrix} 2 0 \\ 3 4 \end{matrix} | - 2 . | \begin{matrix} 2 0 \\ 0 1 \end{matrix} | = 0

ℓ \equiv 3 x + 8 y - 4 = 0

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